DOI: 10.2478/auom-2013-0024
An. Şt. Univ. Ovidius Constanţa
Vol. 21(2),2013, 81–94
Composition hyperrings
Irina Cristea and Sanja Jančić-Rašović
Abstract
In this paper we introduce the notion of composition hyperring. We
show that the composition structure of a composition hyperring is determined by a class of its strong multiendomorphisms. Finally, the three
isomorphism theorems of ring theory are derived in the context of composition hyperrings.
1
Introduction
The hyperrings have appeared as a new class of algebraic hyperstructures
more general than that of hyperfields, introduced by Krasner [9] in the theory
of valued fields. A Krasner hyperring is a nonempty set R endowed with
a hyperoperation (the addition) and a binary operation (the multiplication)
such that (R, +) is a canonical hypergroup, (R, ·) is a semigroup and the
multiplication is distributive with respect to the addition. The theory of these
hyperrings has been developing since the beginning of seventies, thanks to the
contributions of Mittas [14, 15], Krasner [10], Stratigopoulos [20], till nowadays
[2, 3, 5, 8, 13, 17].
Several types of hyperrings have been proposed (for more details see [6, 11,
12, 16, 22] and their references), but the most general one is that introduced by
Spartalis [18], used also in the context of P -hyperrings or (H, R)-hyperrings
[19]. A comprehensive review of hyperrings theory is covered in Nakassis
[16], Vougiouklis [22] and in the book [6] written by Davvaz, Leoreanu-Fotea.
New applications of the theory of hyperrings in number theory and algebraic
geometry can be found in [4, 21].
Key Words: Hyperring; Hyperideal; Multiendomorphism; Composition hyperoperation.
2010 Mathematics Subject Classification: Primary 20N20; Secondary 06E20.
Received: May 2012.
Accepted: June 2012.
81
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Irina Cristea and Sanja Jančić-Rašović
Based on the notion of composition ring introduced by Adler [1], we define
here the concept of composition hyperring, as a quadruple (R, +, ·, ◦) such that
(R, +, ·) is a commutative hyperring in the general sense of Spartalis, and the
composition hyperoperation ◦ is an associative hyperoperation, distributive
to the right side with respect to the addition and multiplication. Many of
the familiar rings of functions are composition rings, where the composition
operation is defined just as the composition between the functions. The idea
to study a similar hyperstructure comes from the properties of the operations
between the polynomials with coefficients in a commutative hyperring, the set
of them forming a hyperring as shown in [7] by Jančić-Rašović.
The rest of the paper is organized as follows. After a short presentation of
the main results from hyperring theory covered in Preliminaries, in Section 3,
we define the notion of composition hyperring, proving that the composition
structure of it is determined by a certain class (φy )y∈R of strong multiendomorphisms of the considered hyperring R. We determine conditions under which
an arbitrary family Ω of multiendomorphims of R generates the class (φy )y∈R
of strong multiendomorphisms of R, and, consequently, the composition hyperoperation on R. In Section 4, using the notion of composition hyperideal of
a composition hyperring, the three isomorphism theorems of ring theory are
derived and discussed in the context of composition hyperrings. We end this
note with some concluding remarks and some open problems.
2
Preliminaries
We recall some definitions concerning hyperrings theory and we fix the notations used in this paper.
A canonical hypergroup is a nonempty set H endowed with an additive
hyperoperation + : H × H −→ P∗ (H), satisfying the following properties:
1. for any x, y, z ∈ H, x + (y + z) = (x + y) + z
2. for any x, y ∈ H, x + y = y + x
3. there exists 0 ∈ H such that 0 + x = x + 0 = x, for any x ∈ H
4. for every x ∈ H, there exists a unique element x0 ∈ H, such that 0 ∈
x + x0 (we write −x instead of x0 and we call it the opposite of x.)
5. z ∈ x + y implies that y ∈ −x + z and x ∈ z − y, that is (H, +) is
reversible.
A multivalued system (R, +, ·) is a hyperring (in the general case of Spartalis [18]), if:
COMPOSITION HYPERRINGS
83
1. (R, +) is a hypergroup
2. (R, ·) is a semihypergroup
3. the multiplication is distributive with respect to the addition, i.e. for all
x, y, z ∈ R, x · (y + z) = x · y + x · z and (x + y) · z = x · z + y · z.
If R is commutative with respect to both addition and multiplication, then it
is called a commutative hyperring.
A particular case of hyperring is that called Krasner hyperring, where
(R, +) is a canonical hypergroup, (R, ·) is a semigroup having 0 as a bilaterally
absorbing element, and the multiplication is distributive with respect to the
addition.
A nonempty subset S of a hyperring R is called a subhyperring of R if,
(S, +) is a subhypergroup of (R, +) and S · S ⊆ S. Moreover, a subhyperring
S of a hyperring R is a hyperideal of R, if r · x ⊆ S and x · r ⊆ S, for all r ∈ R
and x ∈ S.
Suppose now that (R, +, ·) and (T, +0 , ·0 ) are two hyperrings. A map φ :
R −→ T is called a multihomomorphism from R to T if, for all x, y ∈ R, the
following relations hold:
S
1. u∈x+y φ(u) ⊆ φ(x) +0 φ(y)
S
2. u∈x·y φ(u) ⊆ φ(x) ·0 φ(y)
If, in the previous conditions, the equality is valid, then φ is called a strong
multihomomorphism. A multihomomorphism from R to R is called multiendomorphism of R. If φ1 and φ2 are multiendomorphisms
S on a hyperring R,
then their composition φ1 ◦ φ2 defined by (φ1 ◦ φ2 )(x) = a∈φ2 (x) φ1 (a) is also
a multiendomorphism on R.
3
Composition hyperrings
In this section we introduce the notion of composition hyperring, giving several
examples that illustrate the significance of this new hyperstructure. The composition rings constructed by Adler [1] represents a special case of composition
hyperrings.
A composition ring is a commutative ring R with an additional binary
operation ◦ (called composition), satisfying the following properties:
1. (x + y) ◦ z = x ◦ z + y ◦ z
2. (xy) ◦ z = (x ◦ z)(y ◦ z)
3. x ◦ (y ◦ z) = (x ◦ y) ◦ z,
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Irina Cristea and Sanja Jančić-Rašović
for any x, y, z ∈ R. The most significant and natural example of a such ring is
represented by the ring of functions, where the composition operation is just
the composition between two functions: (f ◦ g)(x) = f (g(x)). Extending this
construction to the case of hyperstructures, we obtain the following concept.
Definition 3.1. A composition hyperring is an algebraic structure (R, +, ·, ◦),
where (R, +, ·) is a commutative hyperring and the hyperoperation ◦ satisfies
the following properties, for any x, y, z ∈ R:
1. (x + y) ◦ z = x ◦ z + y ◦ z
2. (x · y) ◦ z = (x ◦ z) · (y ◦ z)
3. x ◦ (y ◦ z) = (x ◦ y) ◦ z.
The binary hyperoperation ◦ having the previous properties is called the composition hyperoperation of the hyperring (R, +, ·).
Definition 3.2. Let (R, +, ·, ◦) be a composition hyperring. An element c ∈ R
is called a constant, if c ◦ x = c, for all x ∈ R. If A is an arbitrary subset of R,
the set of all constants in A is called a foundation of A, denoted by F oundA.
Example 3.3. Let (R, +, ·) be a commutative hyperring. A formal power
series with coefficients in R is an infinite sequence (a0 , a1 , . . . , an , . . .) in which
all ai belong to R. The set of all formal power series with coefficients in R
will be denoted by R[[x]]. Defining the following hyperoperations ⊕ and by
taking:
(a0 , a1 , . . ., an , . . .) ⊕ (b0 , b1 , . . ., bn , . . .)={(c0 , c1 , . . ., cn , . . .) | ck ∈ ak + bk }
and
(a0 , a1 , . . ., an , . . .) (b0 , b1 , . . ., bn , . . .) = {(c0 , c1 , . . ., cn , . . .) | ck ∈
X
ai bj },
i+j=k
then the obtained hyperstructure (R[[x]], ⊕, ) is a hyperring.
Suppose now that the hypergroup (R, +) has one identity, the zero element
0. Let R[x] denote the set of all polynomials (a0 , a1 , . . . , an , . . .) of R[[x]]
having ai = 0 except a finite number of indices i. If 0 + 0 = 0 and a · 0 = 0,
for all a ∈ R, then according to Theorem 3.2. [7], it follows that (R[x], ⊕, )
is a subhyperring of (R[[x]], ⊕, ).
Take f = (a0 , a1 , . . . , an , . . .) ∈ R[x] such that ak = 0, for all k ≥ n + 1,
and take g ∈ R[x]. Define a new hyperoperation by putting:
f ◦ g = a0 ⊕ (a1 g) ⊕ . . . ⊕ (an g n ).
It can be easily verified that (R[x], ⊕, , ◦) is a composition hyperring with
F ound(R) = R.
85
COMPOSITION HYPERRINGS
Example 3.4. If (R, +, ·) is an arbitrary commutative hyperring and ◦ is
defined by r ◦ s = r, for all r, s ∈ R, then (R, +, ·, ◦) is a composition hyperring
with F ound(R) = R.
Example 3.5. Let (R, +, ·) be a commutative ring and RR be the ring of all
functions from R into R. If we define the binary operation ◦ as the composition
of functions, then (RR , +, ·, ◦) becomes a composition ring with F ound(R) =
R.
In the following we propose a method to define the composition structure of
a composition hyperring by a certain class of its strong multiendomorphisms.
Theorem 3.6. Let (R, +, ·, ◦) be a composition hyperring. For any element
y ∈ R, the function Φy : R −→ P∗ (R) defined by Φy (x) = x ◦ y, for all
x ∈ R, is a strong multiendomorphism of the hyperring R. Moreover, if M is
a nonempty subset of R, denote by
[
ΦM (x) =
Φm (x), ∀x ∈ R.
m∈M
Then, for all x, y, z ∈ R, it holds:
ΦΦx (y) (z) =
[
Φx (t).
(1)
t∈Φy (z)
Conversely, if (R, +, ·) is a commutative hyperring and (Φy )y∈R is a family
of its strong multiendomorphisms satisfying equation (1), then, defining the
hyperoperation ◦ by x ◦ y = Φy (x), we obtain that (R, +, ·, ◦) is a composition
hyperring.
Proof. Let (R, +, ·, ◦) be a composition hyperring and let y ∈ R. By the
definition of the function Φy , for all a, b ∈ R, it holds:
[
[
Φy (u) =
u ◦ y = (a + b) ◦ y = a ◦ y + b ◦ y = Φy (a) + Φy (b)
u∈a+b
u∈a+b
and
[
u∈a·b
Φy (u) =
[
u ◦ y = (a · b) ◦ y = (a ◦ y) · (b ◦ y) = Φy (a) · Φy (b).
u∈a·b
Thus, Φy is a strong multiendomorphism of the hyperring (R, +, ·). Moreover,
for all x, y, z ∈ R, it holds:
S
S
ΦΦx (y) (z) = s∈y◦x Φs (z) = s∈y◦x z ◦ s = z ◦ (y ◦ x) = (z ◦ y) ◦ x =
S
S
= t∈z◦y t ◦ x = t∈Φy (z) Φx (t).
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Irina Cristea and Sanja Jančić-Rašović
Suppose now that (Φy )y∈R is a family of strong multiendomorphisms of a
hyperring (R, +, ·) satisfying condition (1). If a hyperoperation ◦ is defined
taking x ◦ y = Φy (x), then it can be easily verified that, for all a, b, x ∈ R, it
holds (a + b) ◦ x = a ◦ x + b ◦ x and (a · b) ◦ x = (a ◦ x) · (b ◦ x).
Besides, using equation (1), we obtain
(a ◦ b) ◦ c =
S
=
S
s∈Φb (a)
u∈Φc (b)
s◦c=
S
Φu (a) =
s∈Φb (a)
Φc (s) = ΦΦc (b) (a)
S
a ◦ u = a ◦ (b ◦ c).
u∈b◦c
Thus, (R, +, ·, ◦) is a composition hyperring and the proof is now complete.
It arises the following question: Can every commutative hyperring
give a composition structure? In this proposal we determine conditions
under which a family of multiendomorphisms of a hyperring (R, +, ·) generates
the class (Φy )y∈R satisfying the conditions of the previous theorem.
Let Ω be a family of multiendomorphisms of a hyperring (R, +, ·). For any
y ∈ R, denote
[
Py =
Φ(y).
Φ∈Ω
The set Py is called the orbit of y. An orbit P is said to be principal if, for all
x ∈ P and Φ1 , Φ2 ∈ Ω, it holds:
Φ1 (x) ∩ Φ2 (x) 6= ∅ =⇒ Φ1 = Φ2 .
Let (R, +, ·) be a commutative hyperring and 0 be an identity element of
the hypergroup (R, +, ·).
Lemma 3.7. Let Ω be a family of strong multiendomorphisms of a hyperring
(R, +, ·), such that:
1. S
Φ1 ◦Φ2 ∈ Ω, for all Φ1 , Φ2 ∈ Ω, where Φ1 ◦Φ2 is defined by: (Φ1 ◦Φ2 )(x) =
v∈Φ2 (x) Φ1 (v).
2. Φ(0) = 0
3. For all x, y ∈ R it holds:
Φ ∈ Ω and x ∈ Φ(y) =⇒ ∃Φ1 ∈ Ω such that y ∈ Φ1 (x).
Then, Ω induces a partition of the set Ω(R) =
S
Φ∈Ω,r∈R
Φ(r) into orbits.
87
COMPOSITION HYPERRINGS
S
Proof. It is clear that Ω(R) = y∈R Py .
First we prove that x ∈ Py implies that Px = Py . Indeed, if x ∈ Py , then
x ∈ Φ(y), for some Φ ∈ Ω. By the third condition of the hypothesis, it follows
that y ∈ Φ1 (x), for some Φ1 ∈ Ω. Thus, if z ∈ Py , then there exists Φ2 ∈ Ω
such that z ∈ Φ2 (y) and so z ∈ (Φ2 ◦ Φ1 )(x). Since Φ2 ◦ Φ1 ∈ Ω, by the
first condition of the hypothesis, we obtain z ∈ Px . Therefore, x ∈ Py implies
Py ⊆ Px . Moreover, if x ∈ Py , then y ∈ Px and consequently Px ⊆ Py , that is
Px = Py .
Thus if Px ∩ Py 6= ∅, then there exists z ∈ Px ∩ Py which implies that
Px = Py = Pz as we have proved before. Thereby, Ω(R) can be partitioned
into the orbits Py , y ∈ R.
Notice that, if the family Ω satisfies the three conditions of the previous
lemma and if Ω has at least two elements, then, for any principal orbit P , it
holds 0 ∈
/ P.
Theorem 3.8. Let Ω be a family of strong multiendomorphisms of a hyperring (R, +, ·) satisfying conditions of Lemma 3.7. Let S be a nonempty set of
principal orbits with 0 ∈
/ S and for each P ∈ S, let ap be an element of P .
For each y ∈ R define the multiendomorphism Φy : R −→ P∗ (R) as follows.
If y is an element of an orbit S
P ∈ S, then Φy = Φ, where Φ is an element of
Ω such that y ∈ Φ(ap ). If y ∈
/ P ∈S P , then Φy = 0.
Then, the family (Φy )y∈R satisfies equation (1), thus it generates a composition hyperoperation on R.
Proof. If P1 , P2 ∈ S and y ∈ P1 ∩ P2 , then, by Lemma 3.7, it follows that
P1 = P2 . Besides, if y ∈ P and y ∈ Φ1 (ap ) ∩ Φ2 (ap ), then Φ1 = Φ2 , because
P is a principal orbit. So the mapping y −→ Φy is well defined.
Obviously, (Φy )y∈R is a family of strong multiendomorphisms of R.
Let x, y, a ∈ R. We will prove the following relation:
[
ΦΦx (y) (a) =
Φx (v).
(2)
v∈Φy (a)
We have to consider
the following situations.
S
1) If x ∈ P ∈S P , then Φx = Φ, where Φ is an element of Ω such that
x ∈ Φ(ap ), with x ∈ P . We have two possibilities:
a) If y ∈ P 0 , for some P 0 ∈ S, then Φy = Φ0 , where Φ0 is an element of Ω
such that y ∈ Φ0 (ap0 ). Thus, Φx (y) = Φ(y) ⊆ (Φ ◦ Φ0 )(ap0 ). Therefore, if
0
z ∈ Φx (y), then z ∈ (Φ ◦ ΦS
)(ap0 ). Since Φ ◦ Φ0 ∈ Ω, it follows that z ∈ P 0
0
and Φz = Φ ◦ Φ . Thus, z∈Φx (y) Φz (a) = (Φ ◦ Φ0 )(a) = (Φx ◦ Φy )(a).
So, the equation (2) holds.
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Irina Cristea and Sanja Jančić-Rašović
S
S
b) Suppose y ∈
/ P ∈S P . Then Φx (y) ∩ P ∈S P = ∅. Indeed, if z ∈
Φx (y) ∩ P 0 , for some P 0 ∈ S, then z ∈ Φx (y) = Φ(y) and so there exists
Φ1 ∈ Ω such that y ∈ Φ1 (z). Since z ∈ P 0 , there exists Φ2 ∈ Ω such that
z ∈ Φ2 (ap0 ). Therefore, y ∈ (Φ1 ◦ Φ2 )(ap0 ), and because Φ1 ◦ Φ2 ∈ Ω, we
obtain y ∈ P 0 , which contradicts the assumption.
S
Therefore, if y ∈
/ P ∈S P
S, then, for all z ∈ Φx (y), it holds Φz = 0 and
so ΦΦx (y) (a) = 0. Also, v∈Φy (a) Φx (v) = Φx (0) = 0.
S
2) Suppose now x ∈
/ P ∈S P . Then Φx = 0 and so Φx (y) = 0. Thus,
ΦΦx (y) (a) = Φ0 (a). S
Notice that 0 ∈
/ P ∈S P . Indeed, S
if 0 ∈ P , for some P ∈ S, then, by
Lemma 3.7, it follows that P = P0 = SΦ∈Ω Φ(0) = 0, i.e. we obtain 0 ∈ S,
contrary to the hypothesis. Thus,
0∈
/ P ∈S P and so Φ0 = 0, i.e. Φ0 (a) =
S
0. So, ΦΦx (y) (a) = 0. Also v∈Φy (a) Φx (v) = 0, since Φx = 0. Thus, the
family (Φy )y∈R satisfies conditions of Theorem 3.6, generating a composition
hyperoperation on R.
Remark 3.9. Let (R, +, ·) be a Krasner hyperring and AutR be the group
of its ordinary automorphisms. If Ω is a subgroup of AutR, then Ω satisfies
conditions of previous theorem. The composition hyperoperation ◦ associated
with the corresponding family (Φy )y∈R is an ordinary operation, since, for all
x, y ∈ R, |x ◦ y| = |Φy (x)| = 1.
Example 3.10. Let (R, +, ·) be the field R of real numbers and A = 2Q =
{2q | q ∈ Q}. Define hyperoperations ⊕A and A on R by: x ⊕A y = xA + yA
and x A y = xAy. It can be easily verified that (R, ⊕A ) is a commutative
hypergroup and (R, A ) is a commutative semihypergroup. Moreover, since,
for all a ∈ A, it holds aA = A, it follows that:
(x ⊕A y) A z
= (xA + yA)Az = (xAz + yAz)A = (xzA + yzA)A =
S
= a∈A (xzAa + yzAa) = xzA + yzA = xzAA + yzAA =
= (x A z) ⊕A (y A z),
for all x, y, z ∈ R. Thus, (R, ⊕A , A ) is a commutative hyperring.
Let us define now two functions f : R −→ P∗ (R) and g : R −→ P∗ (R) by
f (x) = A·x = {2q ·x | q ∈ Q} and g(x) = −A·x = {−2q ·x | q ∈ Q}. Obviously,
f and g are strong multiendomorphisms of (R, ⊕A , A ). Also, f ◦f = g ◦g = f
and f ◦ g = g ◦ f = g. If x ∈ f (y), then x = 2q y, for some q ∈ Q, and so
y = 2−q x ∈ Ax = f (x). Similarly, x ∈ g(y) implies that y ∈ g(x). Obviously
f (0) = 0 and g(0) = 0. Let Ω = {f, g}. It is easy to verify that Ω satisfies
conditions of Lemma 3.7.
COMPOSITION HYPERRINGS
89
Besides, for any y ∈ R, its orbit has the form Py = f (y) ∪ g(y) = {±2q · y |
q ∈ Q}.
If y 6= 0, then Py is a principal orbit, since, for any x ∈ Py , it holds
f (x) ∩ g(x) = ∅, because 2Q x ∩ (−2Q x) = ∅. Thus, by the previous theorem, each family S of principal orbits generates corresponding composition
hyperoperation on R.
S
For instance, if S = {Pn | n ∈ N}, then, for y ∈ Sn∈N Pn and y > 0, we
put Φy = f and, for y < 0, we put Φy = g. If y ∈
/ n∈N Pn , then Φy = 0.
Thus, the corresponding hyperoperation is defined by:

Q

if y ∈ 2Q · N,
2 x
x ◦ y = −2Q x if y ∈ −2Q · N,


0
otherwise.
4
Isomorphism theorems of composition hyperrings
This section deals with the isomorphism theorems for the composition hyperrings. In order to state them, first we introduce the notion of composition
hyperideal and then we construct the quotient composition hyperring.
Throughout this section, (R, +, ·, ◦) is a composition hyperring such that
(R, +) is a canonical hypergroup and x · 0 = 0, for all x ∈ R. Obviously, in a
Krasner hyperring these conditions are satisfied.
Definition 4.1. Let (R, +, ·, ◦) be a composition hyperring and N be a subset
of R. N is called a composition hyperideal of R if the following three conditions
are satisfied:
1. N is a hyperideal of the hyperring R
2. n ◦ r ⊆ N , for all n ∈ N and r ∈ R
3. If r, s, t ∈ R and r − s ∩ N 6= ∅, then t ◦ r − t ◦ s ⊆ N .
Let N be a composition hyperideal of R. Consider on R the following
relation:
xρy ⇐⇒ x + N = y + N.
Obviously, ρ is an equivalence on R and the equivalence class represented by
x is [x]ρ = x + N . Denote by R/N the set of all equivalence classes of the
elements of R with respect to the equivalence relation ρ.
Lemma 4.2. Let R be a composition hyperring and N be a composition hyperideal of R. Defining on the quotient R/N the hyperoperations ⊕, , } as it
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Irina Cristea and Sanja Jančić-Rašović
follows:
(x + N ) ⊕
(y + N )
= {z + N | z ∈ x + y}
(x + N ) (y + N )
= {z + N | z ∈ x · y}
(x + N ) }
(y + N )
= {z + N | z ∈ x ◦ y},
we obtain that (R/N, ⊕, , }) is a composition hyperring, called the quotient
composition hyperring related to the equivalence ρ.
Proof. First we prove that the hyperoperations ⊕, , } are well-defined.
Let x + N = x1 + N and y + N = y1 + N , for x, x1 , y, y1 ∈ R. Set
L = (x + N ) ⊕ (y + N ) = {z + N | z ∈ x + y} and D = (x1 + N ) ⊕ (y1 + N ) =
{z +N | z ∈ x1 +y1 }. If z ∈ x+y, then z +N ⊆ x+y +N = (x+N )+(y +N ) =
(x1 + N ) + (y1 + N ) = x1 + y1 + N . Since z ∈ z + N ⊆ x1 + y1 + N , there
exists z1 ∈ x1 + y1 and n1 ∈ N , such that z ∈ z1 + n1 . It follows that
z + N ⊆ z1 + n1 + N = z1 + N . But z ∈ z1 + n1 , so z1 ∈ z − n1 and then
z1 + N ⊆ z − n1 + N = z + N . Thus, z + N = z1 + N , while z1 ∈ x1 + y1 .
Therefore, L ⊆ D. Similarly one proves that D ⊆ L.
Now set L = (x + N ) (y + N ) = {z + N | z ∈ x · y} and D = (x1 +
N ) (y1 + N ) = {z + N | z ∈ x1 · y1 }. Let z ∈ x · y. Since x ∈ x1 + N
and y ∈ y1 + N , there exist n1 , n2 ∈ N such that z ∈ (x1 + n1 ) · (y1 + n2 ) =
x1 · y1 + n1 · y1 + x1 · n2 + n1 · n2 ⊆ x1 · y1 + N . Thereby, there exist z1 ∈ x1 · y1
and n1 ∈ N such that z ∈ z1 + n1 . It implies that z + N = z1 + N , i.e.
z + N ∈ D. So, L ⊆ D. The converse inclusion can be similarly proved.
Suppose now L = (x + N ) } (y + N ) = {z + N | z ∈ x ◦ y} and D =
(x1 + N ) } (y1 + N ) = {z + N | z ∈ x1 ◦ y1 }. Let z ∈ x ◦ y. Because y ∈ y1 + N ,
we can write y ∈ y1 + n1 , for some n1 ∈ N . It follows that n1 ∈ y − y1 ,
that is y − y1 ∩ N 6= ∅. So x ◦ y − x ◦ y1 ⊆ N , i.e. x ◦ y ⊆ x ◦ y1 + N .
Since x ∈ x1 + N , there exists n2 ∈ N such that x ∈ x1 + n2 and then
x ◦ y ⊆ (x1 + n2 ) ◦ y1 + N = x1 ◦ y1 + n2 ◦ y1 + N ⊆ x1 ◦ y1 + N. Thereby,
if z ∈ x ◦ y, then there exists z1 ∈ x1 ◦ y1 such that z ∈ z1 + N and thus
z + N = z1 + N , which means that L ⊆ D and similarly D ⊆ L.
Finally, it is easy to verify that (R/N, ⊕, , }) is a composition hyperring.
We omit here the classical proof.
Definition 4.3. Let R1 and R2 be composition hyperrings. A mapping f :
R1 −→ R2 is called a strong homomorphism if the following conditions are
satisfied, for all x, y ∈ R1 :
1. f (x + y) = f (x) + f (y)
2. f (x · y) = f (x) · f (y)
91
COMPOSITION HYPERRINGS
3. f (x ◦ y) = f (x) ◦ f (y)
4. f (0) = 0.
A strong homomorphism f is an isomorphism if f is one to one and onto. We
write R1 ∼
= R2 if R1 is isomorphic with R2 .
Notice that, if f is a strong homomorphism from R1 into R2 , then, for all
x ∈ R1 , it holds f (−x) = −f (x). Indeed, since 0 ∈ x − x, it follows that
0 = f (0) ∈ f (x) + f (−x), so f (−x) = −f (x).
If f is a strong homomorphism from R1 into R2 , then the kernel of f is
the set Kerf = {x ∈ R1 | f (x) = 0}. Obviously, Kerf is a hyperideal of
(R1 , +, ·), but generally it is not a composition hyperideal.
In the following, we will state and prove the isomorphism theorems for
composition hyperrings. Note that, for the first theorem we need Kerf to be
a composition hyperideal.
Theorem 4.4. Let R1 and R2 be composition hyperrings. If f is a strong homomorphism from R1 into R2 with the kernel K such that K is a composition
hyperideal of R1 , then R1 /K ∼
= Imf .
Proof. Define φ : R1 /K −→ Imf by φ(x + K) = f (x), for all x ∈ R1 . First
we prove that φ is well-defined. Suppose that x + K = y + K. Then, there
exists z ∈ K such that x ∈ y + z. It follows that z ∈ (−y + x) ∩ K, that is
0 = f (z) ∈ f (x) − f (y). Thus, f (x) = f (y). Obviously, f is onto. It remains to
show that φ is one to one. Suppose φ(x + K) = φ(y + K). Then f (x) = f (y),
which means that 0 ∈ f (x − y). Thus, there exists z ∈ x − y such that
z ∈ K = Kerf and so, x ∈ z + y ⊆ K + y which implies that x + K = y + K.
Thereby φ is a bijection.
Moreover, φ is a strong homomorphism, because
φ((x + K) ⊕ (y + K))
= φ({z + K | z ∈ x + y}) = {f (z) | z ∈ x + y} =
= f (x + y) = f (x) + f (y) = φ(x + K) + φ(y + K).
Similarly,
φ((x + K) (y + K))
= φ(x + K) · φ(y + K),
φ((x + K) } (y + K))
= φ(x + K) ◦ φ(y + K)
and φ(K) = f (0) = 0.
Theorem 4.5. If A and B are composition hyperideals of a composition hyperring R, then A/(A ∩ B) ∼
= (A + B)/B.
92
Irina Cristea and Sanja Jančić-Rašović
Proof. Clearly A ∩ B is a composition hyperideal of A (as the intersection
between composition hyperideals) and A + B is a subhyperring of (R, +, ·). If
x, y ∈ A + B, then there exist a, a1 ∈ A and b, b1 ∈ B such
S that x ∈ a + b and
yS∈ a1 + b1 , and therefore
x
◦
y
⊆
(a
+
b)
◦
(a
+
b
)
=
1
1
s∈a1 +b1 (a + b) ◦ s =
S
0
0
(a◦s+b◦s)
⊆
a
+b
=
A+B.
Thus
A+B
is a composition
0
0
s∈a1 +b1
a ∈A,b ∈B
hyperring and, since B is a composition hyperideal of A + B, it follows that
(A + B)/B is well defined.
Let us take f : A −→ (A + B)/B by f (a) = a + B. It is easy to verify that
f is a strong homomorphism.
We prove that f is onto. Let y + B ∈ (A + B)/B, with y ∈ a + b, for some
a ∈ A and b ∈ B. Then a ∈ y − b, that is a ∈ y + B. Thus a + B = y + B, so
f (a) = y + B.
Besides, for any a ∈ A, it holds:
a ∈ Kerf ⇐⇒ f (a) = B ⇐⇒ a + B = B ⇐⇒ a ∈ A ∩ B.
Thereby Kerf = A ∩ B, and by Theorem 4.4, we get the isomorphism A/(A ∩
B) ∼
= (A + B)/B.
Theorem 4.6. If A and B are composition hyperideals of a composition hyperring R such that A ⊆ B, then B/A is a composition hyperideal of R/A and
(R/A)/(B/A) ∼
= R/B.
Proof. As in the previous two theorems, one can verify that B/A is a composition hyperideal of R/A and that the application f : R/A −→ R/B, defined
by f (x + A) = x + B, is a strong homomorphism of R/A onto R/B with
Kerf = B/A.
5
Conclusions and future work
The notion of hyperring is a natural generalization of that of ring, many properties of rings have been transferred to the case of hyperrings. This paper is a
contribution to the development of the theoretical background of hyperrings
starting from rings. The notion of composition ring introduced in 1962 [1] has
been extended to that of composition hyperring, i.e. a hyperring (R, +, ·, ◦)
with a new hyperoperation ◦, called composition, which is associative and distributive with respect to the addition and multiplication of the hyperring. It
is shown that the composition structure of R can be determined by a certain
class of multiendomorphisms of R. The three isomorphism theorems have been
proved for this class of hyperstructures.
This research could be continued further, for instance the theory of hyperideals (maximal or prime hyperideals) could be developed in this context, or
to study the composition near-hyperrings, starting from near-rings.
COMPOSITION HYPERRINGS
93
Acknowledgements. The first author was partially supported by “Agencija za raziskovalno dejavnost Republike Slovenije, program P 1 − 0285”.
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Irina CRISTEA,
Centre for Systems and Information Technologies,
University of Nova Gorica
SI-5000, Vipavska 13, Nova Gorica, Slovenia.
Email: irinacri@yahoo.co.uk
Sanja JANČIĆ-RAŠOVIĆ,
Department of Mathematics, Faculty of Natural Science and Mathematics,
University of Montenegro,
Dzordza Vasingtona bb, 81000 Podgorica, Montenegro.
Email: sabu@t-com.me